What are all the possible rational zeros for #y=x^4-x^3-4x^2+2x+4# and how do you find all zeros?

1 Answer
Apr 5, 2018

Answer:

The zeros for this function are at #x=-sqrt(2)#, #x=-1#, #x=sqrt(2)#, and , #x=2#.

Explanation:

The possible rational zeros are the factors of the coefficients of the first and last term in standard form. The factors of 4 are #+-1#, #+-2#, and #+-4#. We note that when #x=2#, #y=0#. Therefore, #(x-2)# must be a factor of #y#. We can use synthetic division to prove this, but synthetic division is difficult to format properly in Socratic, so let's just write

#x^4-x^3-4x^2+2x+4#

#=x^4-2x^3+x^3-2x^2-2x^2+4x-2x+4#

#=x^3(x-2)+x^2(x-2)-2x(x-2)-2(x-2)#

#=(x-2)(x^3+x^2-2x-2)#

Now we note that #x=-1# is a zero of #x^3+x^2-2x-2# so #x+1# must be a factor of #x^3+x^2-2x-2#.

#(x-2)(x^3+x^2-2x-2)#

#=(x-2)[x^2(x+1)-2(x+1)]#

#=(x-2)(x+1)(x^2-2)#

Now we can factor #x^2-2# as the difference of two squares.

#(x-2)(x+1)(x^2-2)=(x-2)(x+1)(x-sqrt(2))(x+sqrt(2))#

We can see that this expression evaluates to zero when #x=-sqrt(2)#, #x=-1#, #x=sqrt(2)#, and , #x=2#.